删除javascript中对象数组的重复对

Question

Fiddle Example :

I have an array of objects like this:

var bigarr = 
[ 
  [ { name: 'ABC',id: 391},{ name: 'XYZ',id: 545}],
  [ { name: 'EFG',id: 390},{ name: 'XYZ',id: 545}],
  [ { name: 'XYZ',id: 545},{ name: 'ABC',id: 391}],
  [ { name: 'ABC',id: 391},{ name: 'XYZ',id: 545}]
];

How can I remove any of the duplicated pair of objects that have id 545 and 391 to reduce the array to this:

var newbigarr = 
[ 
  [ { name: 'ABC',id: 391},{ name: 'XYZ',id: 545}],
  [ { name: 'EFG',id: 390},{ name: 'XYZ',id: 545}]
];

I have thought of filtering out the duplicated pairs by making a new list of array:

[{391: 391,545: 545},{390: 390,545: 545}]

and then iterating over it and bigarr to build the newbigarr , but my code isn't even able to create that list to begin with.

var test_id = [];
for(var i = 0;i < bigarr.length;i++)
{
   var value_obj = {};
   for(var j in bigarr[i])
   {
     var value = bigarr[i][j]["id"];  
     value_obj[value] = value;
   }
   test_id.push(value_obj);
}

console.log(test_id);

I'm using lodash,so any solution involving lodash is welcomed.

Answer 1

If you are using underscore, you could do something like this:

var bigarr = 
[ 
  [ { name: 'ABC',id: 391},{ name: 'XYZ',id: 545}],
  [ { name: 'EFG',id: 390},{ name: 'XYZ',id: 545}],
  [ { name: 'XYZ',id: 545},{ name: 'ABC',id: 391}],
  [ { name: 'ABC',id: 391},{ name: 'XYZ',id: 545}]
];

console.log(_.uniq(bigarr,function(item){
    return [
        [item[0].id+item[0].name],
        [item[1].id+item[1].name]
    ].sort().toString()
}))

Edit: this only works if you always have one pair of objects with an id and a string name. Deepened objects or other keys are not checked

Answer 2

JSFIDDLE

If your using pure js, you can do this something like this just using iteration and a secondary array to keep track of what you've seen:

var bigarr = 
[ 
  [ { name: 'ABC',id: 391},{ name: 'XYZ',id: 545}],
  [ { name: 'EFG',id: 390},{ name: 'XYZ',id: 545}],
  [ { name: 'XYZ',id: 545},{ name: 'ABC',id: 391}],
  [ { name: 'ABC',id: 391},{ name: 'XYZ',id: 545}]
];

var seen = [];
var final = [];

for(i = 0; i < bigarr.length; i++){
    var unseen = true;
    for(j = 0;j<seen.length;j++){
        if((seen[j][0] == bigarr[i][0].id && seen[j][1] == bigarr[i][1].id) || (seen[j][0] == bigarr[i][1].id && seen[j][1] == bigarr[i][0].id)
          ){
            unseen = false;
        }
    }
    if(unseen){
        final.push(bigarr[i]);
        seen.push([bigarr[i][0].id, bigarr[i][1].id]);
    }
}

Answer 3

Using Lo-Dash. JSFiddle

var bigarr = 
[ 
  [ { name: 'ABC',id: 391},{ name: 'XYZ',id: 545}],
  [ { name: 'EFG',id: 390},{ name: 'XYZ',id: 545}],
  [ { name: 'XYZ',id: 545},{ name: 'ABC',id: 391}],
  [ { name: 'ABC',id: 391},{ name: 'XYZ',id: 545}]
];

var seenIds = {};
var nodup = _.filter(bigarr, function(pair) {
    var ids = [pair[0].id, pair[1].id].sort();
    if(seenIds[ids]) {
        return false;
    } else {
        seenIds[ids] = true;
        return true;
    }
});

Answer 4

One line solution with lo-dash ( jsFiddle ):

var bigarr = 
[ 
  [ { name: 'ABC',id: 391},{ name: 'XYZ',id: 545}],
  [ { name: 'EFG',id: 390},{ name: 'XYZ',id: 545}],
  [ { name: 'XYZ',id: 545},{ name: 'ABC',id: 391}],
  [ { name: 'ABC',id: 391},{ name: 'XYZ',id: 545}]
];  

_.uniq(bigarr, function(a) { return _.pluck(a, 'id').sort() + '' });